Solutions of $n^{n}=2n$?

Question

What is the best way to find solutions to $n^{n}$ and $2n$? Someone suggested the Lambert W function, but I cannot find a way to set it up in a way to use that. Any other suggestions?

Meaning, what are the solutions for $n$ in

$$n^n=2n$$

Answer 1

$n = 2$ is a straighforwardly verifiable solution.

By inspection of the graph of $y = n^n - 2n$, there's another zero in the vicinity of$~0.34...$ that can be found to arbitrary precision numerically.

By noting that $\frac{d(n^n - 2n)}{dn} = n^n(\ln(n)+1) - 2$, the function has only one global minima for $n > 0$ and is increasing elsewhere. Therefore these are the only two zeros.

Answer 2

to make calculation easier...we can start from $$n^n=2n$$ $$\implies n^{n-1}=2$$ $$\implies n^{n-1}-2=0$$ Now let's assume $$f(n)=n^{n-1}-2$$ Let's find the root using Newton's method of calculating numerical root of an equation. the rule is, $$ x_{n+1}=x_n-\dfrac{f(x)}{f'(x)}$$ so,we will need $f'(n)$.So,let's calculate it first. here,if you calculate the derivative you will find, $$f'(n)=n^{n-1}\ln 10(\dfrac{n-1}{n \ln 10}+\log n)$$ So,according to Newton's method $$n_{r+1}=n_r-\dfrac{n_r^{n_r-1}-2}{n_r^{n_r-1}\ln 10(\dfrac{n_r-1}{n_r \ln 10}+\log n_r)}$$

So,let's start our first assumption $$n_o=0.5$$ and continuing the iteration process,we will get $$0.25535...\\0.319893...\\0.344004...\\0.346305...\\0.3463233...\\0.3463233$$ From here we can conclude that,one root is $n \approx 0.346$

Again if we start the iteration process taking the first assumption $n_o=2.5$ we get, $$2.17418...\\2.02494...\\2.00055...\\2.0000002...\\2$$ from here we can conclude that another root is $n=2$

Here, $$0.346^{0.346-1}\approx 2$$ again ,$$2^{2-1}=2$$

Hence,yes our calculated roots are correct. Now,if you want to feel the Lambert W function here, then just do a little work... $$n^n=2n$$ $$\implies n^{n-1}=2$$ $$\text{so,}~n=W(2)=\Omega$$ Now,we can write, $$\Omega^{\Omega-1}-2=0$$ Now,continue the same process as mentioned before.